Let's solve each part step by step, with **explanations**, and **sketching guidelines**. Where possible, I will suggest explicit algebraic examples to make the sketches easier. --- ### (a) The function \( f \) has **one inflection point** but **no local extrema**. #### **Step-by-step Solution** **1. No local extrema** means: - The function does **not** have any critical points where \( f'(x)=0 \) and the derivative changes sign. - So, \( f'(x) \) must **never be zero** (or, if zero, the sign does not change). **2. One inflection point** means: - The function changes concavity **exactly once**. **3. The simplest such function is a cubic with no real critical points.** #### **Example Construction** - A cubic \( f(x) = x^3 + bx + c \) will have 1 inflection point at \( x=0 \). - Its derivative: \( f'(x) = 3x^2 + b \). - For **no extrema**, \( f'(x) \) must not have real roots: \( 3x^2 + b \ne 0 \) for all \( x \). So, \( b > 0 \) or \( b < 0 \); \( 3x^2 + 1 > 0 \) for all real \( x \). **Let’s use:** \( f(x) = x^3 + x \) - \( f'(x) = 3x^2 + 1 \) (always positive) - \( f''(x) = 6x \), which is zero at \( x=0 \) (**inflection point**). **SKETCH:** - Starts from bottom left, steepens, passes through inflection at \( x=0 \), continues steepening to top right. - No peaks or valleys. --- ### (b) The function \( f \) has **three real zeros** and **exactly two local minima**. #### **Step-by-step Solution** **1. Three real zeros:** - Function crosses the x-axis three times. - Usually, the **minimum degree** for this is degree 3 (cubic), but a cubic can have at most **one minimum** and **one maximum**. - So, **we need degree at least 4**. **2. Two local minima:** - So, there must be **at least two “valleys”** in the graph. **3. Therefore, the most straightforward is a **quartic** (degree 4) with two minima and one maximum.** #### **Example Construction** - A “W” shaped quartic can have this property. - General form: \( f(x) = ax^4 + bx^3 + cx^2 + dx + e \) **Let’s try:** \( f(x) = x^4 - 2x^2 + x \) Let’s check: 1. **Zeros:** Set \( x^4 - 2x^2 + x = 0 \) - Try \( x = 0 \): 0 - Try \( x = 1 \): \( 1 - 2 + 1 = 0 \) - Try \( x = -1 \): \( 1 - 2 + (-1) = -2 \) - Try \( x = 2 \): \( 16 - 8 + 2 = 10 \), but not zero. - Not easy to factor, but you can check numerically that it has three real zeros. 2. **Critical points:** Set \( f'(x) = 4x^3 - 4x + 1 \) - This is a cubic, so it can have up to three real roots. - For suitable coefficients, you can get two minima and one maximum. **Or, sketch a function like**: - Downward parabola between two upward parabolas, so the function dips down twice and up once. **SKETCH:** - Starts high left, dips to first minimum, rises to maximum (crossing x-axis), dips to second minimum, rises again. - Crosses x-axis three times. - Minima at the dips, maximum in the middle. --- ### (c) The function \( f \) satisfies: Let's rewrite the conditions more clearly (the notation above is a bit jumbled): - For \( x < -1 \): \( f' < 0 \), \( f'' < 0 \) - For \( -1 < x < 2 \): \( f' > 0 \), \( f'' > 0 \) - For \( 2 < x < 6 \): \( f' > 0 \), \( f'' < 0 \) - For \( x > 6 \): \( f' < 0 \), \( f'' < 0 \) #### **Step-by-step Solution** Let’s interpret this **piece-by-piece**: 1. **For \( x < -1 \):** - **Decreasing** and **concave down**. 2. **At \( x = -1 \):** - \( f' \) switches from negative to positive (minimum). - \( f'' \) switches from negative to positive (inflection point? Possibly, but depends on continuity). 3. **For \( -1 < x < 2 \):** - **Increasing** and **concave up**. 4. **At \( x = 2 \):** - \( f'' \) switches from positive to negative (inflection point). 5. **For \( 2 < x < 6 \):** - **Increasing** and **concave down**. 6. **At \( x = 6 \):** - \( f' \) switches from positive to negative (maximum). - \( f'' \) remains negative. 7. **For \( x > 6 \):** - **Decreasing** and **concave down**. #### **Key features:** - **Local minimum at \( x = -1 \)** - **Inflection point at \( x = 2 \)** - **Local maximum at \( x = 6 \)** - **Concave down at ends, concave up in between -1 and 2** #### **Possible explicit function:** Let’s try to create a **quintic polynomial**, since we need up to three critical points (min, inflection, max): General form: \( f(x) = a(x + 1)^2(x - 2)(x - 6)^2 \) Let’s check the shape: - The squared factors at \( x = -1 \) and \( x = 6 \) mean horizontal tangents (possible extrema). - The factor \( (x-2) \) means it will be zero at \( x=2 \), and will change concavity if constructed right. But for more precise control, let's consider a generic example for the sketch: **SKETCH:** - Decreasing and concave down for \( x < -1 \). - At \( x = -1 \): minimum, starts increasing and concave up. - At \( x = 2 \): inflection, still increasing but now concave down. - At \( x = 6 \): maximum, after which decreasing and concave down. **Shape:** - Looks like a soft "S" that rises from a left minimum, inflects to steeper, then peaks at a max, then falls. --- ## **Summary Table** | Part | Properties | Example Function | Key Features / Sketch | |------|------------|------------------|----------------------| | (a) | One inflection, no extrema | \( f(x) = x^3 + x \) | S-shaped, always increasing, inflection at 0 | | (b) | 3 real zeros, 2 local minima | \( f(x) = x^4 - 2x^2 + x \) | "W" shape, 2 valleys (minima), 1 peak (maximum), crosses x-axis 3 times | | (c) | See above | See above | Minimum at \( x=-1 \), inflection at \( x=2 \), maximum at \( x=6 \); decreasing left, increasing through, then decreasing right; concave down at ends, up in middle | --- If you need help with the **algebraic construction** or would like to see **explicit plots**, let me know!